I know that for any Lebesgue integrable function $f:X\to\mathbb{C}$, or $f:X\to\mathbb{R}$, where $X$ is a set of finite measure such that $X=\bigcup_n A_n$, $\forall i\ne j\quad A_i\cap A_j=\emptyset$, the following equality holds$$\int_X fd\mu=\sum_n\int_{A_n}fd\mu$$where the existence of the first integral implies the existence of the integrals in the second member of the equality.
Then I read that it also holds if $X$ is not of finite measure, but can be represented as the union of countably many sets of finite measure such that $X=\bigcup_{k=1}^\infty X_k$ and $X_1\subset X_2\subset...$. In that case, the integral is defined by $\int_X fd\mu:=\lim_{k\to\infty}\int_{X_k} fd\mu$.
I wonder how the equality above can be generalised in such a case. I guess that a way to prove it might be taking limits in the equality above, which is valid for any $X_k$: $\int_{X_k} fd\mu=\sum_n\int_{A_n\cap X_k}fd\mu$. I would understand why it holds if I could see that $\int_X fd\mu:=\lim_{k}\int_{X_k}fd\mu=\lim_k\sum_n\int_{A_n\cap X_k}fd\mu$ is equal to $\sum_n\lim_k\int_{A_n\cap X_k}fd\mu=\sum_n\int_{A_n}fd\mu$, but I do not know whether and why it would be possible to "move the limit(s) across the $\sum$". I heartily thank you for any help!
EDIT: I have realised that it may well be necessary that for any $n$ and $k$ $A_n\cap X_k$ be measurable. The text does not specify the condition, but I think it's quite obvious.